# Integrate a top form over a surface without partition of unity

Suppose we are given a compact Riemann surface $$M$$, an open cover $$\mathscr{U}=\{U_1,U_2,\dots\}$$ of $$M$$, charts $$\{(U_1,\phi_1),(U_2,\phi_2),\dots\}$$, holomorphic coordinates, $$\phi_m:p\in U_m\mapsto z_m$$, holomorphic transition functions $$z_m=f_{mn}(z_n)$$ on patch overlaps $$U_m\cap U_n$$ (with appropriate cocycle relations, $$f_{mn}\circ f_{n\ell}\circ f_{\ell m}=1$$ on triple overlaps $$U_m\cap U_n\cap U_\ell$$), and a globally defined top form, $$\omega=\omega(z,\bar{z})dz\wedge d\bar{z}$$.

I want to integrate $$\omega$$ over $$M$$, $$I= \int_M\omega,$$ using the above data (without using a partition of unity), and I'm wondering whether my reasoning is correct. In particular, suppose we construct non-overlapping sets $$\{V_1,V_2,\dots\}$$ such as those shown in the figure: My question is whether I can compute $$I$$ via a sum of integrals over the $$\{V_1,V_2,\dots\}$$ (to guarantee no overcounting) without using a partition of unity; i.e., is it true that: \begin{aligned} I &= \sum_m\int_{V_m}\omega\\ &=\sum_m\int_{V_m}\omega_m(z_m,\bar{z}_m)dz_m\wedge d\bar{z}_m \end{aligned} In the examples I have considered this gives the right answer but I would like to know if it is true in general. Thanks!

Yes, from a computational perspective that works fine, assuming that one can describe the decomposition $$\{V_1,V_2,\ldots\}$$ with sufficient rigor (in particular, their boundaries should have measure zero).

The proof would go something like this.

Define a new partition $$\{W_k\}$$ whose elements are obtained by taking all possible finite intersections of elements of the $$\{U_i\}$$ covering and elements of the $$\{V_j\}$$ decomposition. For example, in your picture one of the $$W$$'s would be $$U_n \cap U_l \cap V_m$$ which is the tiny quadrilateral in your $$V$$ diagram having two straight sides on the boundary of $$V_m$$, one curved side on the boundary of $$U_n$$, and one curves side on the boundary of $$U_l$$.

First one proves that $$I = \int_M \omega = \sum_k \int_{W_k} \omega$$. One has $$\int_M \omega = \sum_i \int_{U_i} \phi_i \omega$$ by definition. Then one decomposes $$U_i$$ into a disjoint union of $$W_k$$'s, and rewrites $$\int_{U_i} \phi_i \omega$$ as a sum over the terms of that disjoint union. Then for each $$k$$ one collects the $$W_k$$ terms and takes their sum, using the partition of unity property to get a sum equal to $$\int_{W_k} \omega$$.

Added: By request, here are more details in the case of $$W_k = U_n \cap U_l \cap V_m$$. The terms that one collects to get $$\int_{W_k} \omega$$ are: $$\int_{W_k} \phi_n \omega + \int_{W_k} \phi_l \omega + \int_{W_k} \phi_m \omega = \int_{W_k} (\phi_n+\phi_l+\phi_m) \omega = \int_{W_k} 1 \cdot \omega = \int_{W_k} \omega$$

Next one simply collects the $$\int_{W_k} \omega$$ terms in a different way to get $$\sum_j \int_{V_j} \omega$$.

• excellent!! thank you – Wakabaloola Mar 6 at 15:25
• it would be helpful (also for future readers) if you could add a bit more detail on constructing the integral, $\int_{W_k}\omega$, over the specific quadrilateral you mentioned in the example, in particular how the partition of unity enters and what precise terms are contained in the sum for the given $k$ – Wakabaloola Mar 6 at 15:44
• I added a few more details. – Lee Mosher Mar 6 at 16:31
• that's now crystal clear and very helpful -- many thanks again – Wakabaloola Mar 6 at 16:33